Question

The given figure shows image formation by a lens. Analyse the figure and answer the following questions : 


What is the type of lens used for image formation in the given ray diagram ? 
If the real image is formed at a distance of \( 30 \text{ cm} \) from the lens and the size of image is twice the size of the object, then where was the object placed ? 
 

Hint:

Always use sign convention: Real images in lenses have positive \( v \), and object distance \( u \) is always negative.

Answer 1

Step 1: Analyzing the Diagram (a):
The ray diagram shows a lens where the parallel incident rays diverge after passing through it. A lens that diverges light rays is a concave lens (diverging lens). It forms a virtual, erect, and diminished image.
Step 2: Key Formula or Approach for (b):
Part (b) is a numerical problem. Note that a concave lens cannot form a real image. Therefore, part (b) must refer to a convex lens which can form real images.
Given:
- Image is real, so it is on the opposite side: \( v = +30 \text{ cm} \).
- Size is twice: Magnification \( |m| = 2 \).
- Since the image is real, it must be inverted, so \( m = -2 \).
Step 3: Detailed Explanation for (b):
Magnification formula for lens: \( m = \frac{v}{u} \)
\[ -2 = \frac{30}{u} \]
\[ u = \frac{30}{-2} = -15 \text{ cm} \]
Step 4: Final Answer:
(a) The lens in the diagram is a concave lens.
(b) The object was placed at a distance of \( 15 \text{ cm} \) in front of the lens.